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babunello [35]
3 years ago
14

A 1.5m wire carries a 6 A current when a potential difference of 68 V is applied. What is the resistance of the wire?

Physics
1 answer:
ANTONII [103]3 years ago
6 0

Answer:

11.3 \Omega

Explanation:

We can find the resistance of the wire by using Ohm's law:

V=RI

where

V is the voltage applied

R is the resistance

I is the current

In this problem, we know I = 6 A and V = 68 V, so we can re-arrange the equation to find the resistance of the wire:

R=\frac{V}{I}=\frac{68 V}{6 A}=11.3 \Omega

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Ierofanga [76]

\sqrt{2}Answer:

KE2 = 2 KE1

1/2 M V2^2 = 2 * (1/2 M V1^2)

V2^2 = 2 V1^2

V2 = \sqrt{2} V1

Since momentum = M V  the momentum increases by \sqrt{2}

8 0
3 years ago
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5. A person fishing from a pier observes that four wave crests pass by in 7.0 s and estimates that the distance between two succ
TiliK225 [7]

Answer:

v= 1.71 m/s

Explanation:

Given that

Distance between two successive crests = 4.0 m

 λ = 4 m

T= 7 sec

T is the time between 3 waves.

3 waves = 7 sec

1 wave = 7 /3 sec

So t= 7/3 s

We know that frequency f

f= 1/t= 3/7 Hz

Lets take speed of the wave is v

v= f λ

f=frequency

λ=wavelength

v= 3/7 x 4 = 12 /7

v= 1.71 m/s

3 0
3 years ago
A 65.0-Ω resistor is connected to the terminals of a battery whose emf is 12.0 V and whose internal resistance is 0.5 Ω. Calcula
Luda [366]

Answer:

a) 0.1832 A

b) 11.91 Volts

c) 2.18 Watt , 0.0168 Watt

Explanation:

(a)

R = external resistor connected to the terminals of the battery = 65 Ω

E = Emf of the battery = 12.0 Volts

r = internal resistance of the battery = 0.5 Ω

i = current flowing in the circuit

Using ohm's law

E = i (R + r)

12 = i (65 + 0.5)

i = 0.1832 A

(b)

Terminal voltage is given as

V_{ab} = i R

V_{ab} = (0.1832) (65)

V_{ab} = 11.91 Volts

(c)

Power dissipated in the resister R is given as

P_{R} = i²R

P_{R} = (0.1832)²(65)

P_{R} = 2.18 Watt

Power dissipated in the internal resistance is given as

P_{r} = i²r

P_{r} = (0.1832)²(0.5)

P_{r} = 0.0168 Watt

5 0
3 years ago
An assault rifle fires an eight-shot burst in 0.40 s. Each bullet has a mass of 7.5 g and a speed of 300 m/s as it leaves the gu
myrzilka [38]

Answer:

The average recoil force on the gun during that 0.40 s burst is 45 N.

Explanation:

Mass of each bullet, m = 7.5 g = 0.0075 kg

Speed of the bullet, v = 300 m/s

Time, t = 0.4 s

The change in momentum of an object is equal to impulse delivered. So,

F\times t=mv\\\\F=\dfrac{mv}{t}

For 8 shot burst, average recoil force on the gun is :

F=\dfrac{8mv}{t}\\\\F=\dfrac{7.5}{1000}\cdot\dfrac{300}{0.4}\cdot8\\\\F=45\ N

So, the average recoil force on the gun during that 0.40 s burst is 45 N.

5 0
3 years ago
Describe and name the different types of collision. In which are the linear momentum and kinetic energy conserved
ipn [44]

Answer:

1. Elastic collision

2. Inelastic collision    

Explanation:

Elastic collision: collision is said to be elastic if total kinetic energy is not conserved and if there is a rebound after collision

the collision is described by the equation bellow

m1U1+ m2U2= m1V1+m2V2

Inelastic collision: this type of collision occurs when the total kinetic energy of a body is conserved or when the bodies sticks together and move with a common velocity

the collision is described by the equation bellow

m1U1+ m2U2= V(m1+m2)

8 0
3 years ago
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